Question

A 85^B atom (mass = 8.0246 amu) decays into a 84^Be atom (mass = 8.0053 amu) by loss of a β+ particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?

a) 7.45 MeV
b) 5.25 MeV
c) 3.35 MeV
d) 2.10 MeV

Answer 1

The energy produced by the decay of a 85^B atom into a 84^Be atom by either loss of a β+ particle or by electron capture is 9.4047 MeV.

Step-by-step explanation:

The reaction in question involves the decay of a 85B atom into a 84Be atom by either loss of a β+ particle or by electron capture. To calculate the energy produced by this reaction, we can use the mass-energy equivalence formula, E = Δm * c^2, where Δm is the change in mass and c is the speed of light. The change in mass can be calculated by subtracting the mass of the products from the mass of the reactant.

Let's calculate the change in mass:

Change in mass = mass of 85B - (mass of 84Be + mass of β+ particle) = 8.0246 amu - (8.0053 amu + 0.00055 amu) = 0.01875 amu

Now, let's calculate the energy produced:

Energy = Δm * c^2 = 0.01875 amu * (9.314 * 10^8 m/s)^2 = 1.5096 * 10^-6 J

Converting this energy to electron volts, we can use the conversion factor: 1 eV = 1.602 * 10^-19 J

Energy in electron volts = (1.5096 * 10^-6 J) / (1.602 * 10^-19 J/eV) = 9.4047 * 10^12 eV

Finally, to express the energy in millions of electron volts (MeV), we divide by 10^6:

Energy in MeV = (9.4047 * 10^12 eV) / 10^6 = 9.4047 MeV

Therefore, the correct answer is a) 9.4047 MeV.